# Prove that if a sequence $x_n$ tends to infinity, then $\frac{1}{x_n}$ converges to zero.

Question: $$(x_n)_{n=1}^\infty$$ is a sequence with $$x_n\neq0$$ for all $$n$$, also let $$x_n$$ tend to infinity. Let $$(y_n)_{n=1}^\infty$$ be defined by $$y_n=\frac{1}{x_n}$$, show that this converges to zero.

Definition for tending to infinity:$$\forall K \in \mathbb{R} \exists N\in \mathbb{N} \forall n \in \mathbb{N} ,n>N:x_n>K$$

Definition for convergence to zero:$$\forall \varepsilon > 0 \exists N\in \mathbb{N} \forall n \in \mathbb{N} ,n>N:∣x_n∣<\varepsilon$$

Idea: I know that $$x_n$$ tends to infinity since this is assumed. Then it must satisfy the defintion, therefore there must exist an $$N$$ which satisfies the defintion which i will call $$N_x$$ to prevent confusion.

To prove that $$y_n$$ converges to zero then I should choose $$N_y=\lceil\frac{1}{N_x}\rceil$$

Proof (attempt): Given $$\varepsilon>0$$ choose $$N_y=\lceil\frac{1}{N_x}\rceil$$. Given $$n \in \mathbb{N}$$, $$n>N$$ we have $$∣y_n∣=∣\frac{1}{x_n}∣\leq \frac{1}{N_x}\leq\varepsilon$$

Not entirely sure my selection for $$N_y$$ is correct, I understand why the statement is true however the proving put I am struggling.

• For any $\varepsilon >0$ there exists $M \in \Bbb{N}$ such that $1/M< \varepsilon$ solves this, as $1/n$ is obviously decreasing – B.Swan May 11 at 20:53

Take $$\varepsilon>0$$. Since $$\lim_{n\to\infty}x_n=\infty$$, there is some $$N\in\mathbb N$$ such that $$n\geqslant N\implies x_n>\frac1\varepsilon$$. But then $$n\geqslant N\implies 0<\frac1{x_n}<\varepsilon$$. In particular, $$\left\lvert\frac1{x_n}\right\rvert<\varepsilon$$.

• Hi thank you for your reply, would you mind clarifying how it implies that $x_n>\frac{1}{\varepsilon}$? – ViB May 11 at 21:17
• I am using the definition of $\lim_{n\to\infty}x_n=\infty$, taking $K=\frac1\varepsilon$. Is it clear now? – José Carlos Santos May 11 at 21:21

Your intuition is right: the definition of $$N_y$$ is a bit off.

When you choose $$N_x$$, this value is for a particular $$K$$, from the "For all $$K$$ there exists $$N\in\mathbb{N}$$" part of the first statement. As $$K$$ increases without bound, so will $$N_x$$. Here, I think you're choosing $$N_y$$ with the goal of showing that $$y_n$$ is sufficiently small for all $$n\geq N_y$$. But if you let $$K\rightarrow\infty$$, choose a corresponding $$N_x$$, and then let $$N_y=\lceil\tfrac{1}{N_x}\rceil$$, this will result in $$N_x$$ shooting off to $$\infty$$ but $$N_y$$ will end up being $$1$$ all the time. Your goal will then be proving that $$y_n$$ is sufficiently small for all $$n\geq N_y=1$$, which won't be true.

For this problem, you've been given a relationship between the values of $$\{x_n\}$$ and $$\{y_n\}$$, so you should use this relationship to somehow connect the formal statements of "$$x_n\rightarrow\infty$$" and "$$y_n\rightarrow 0$$". Let's restate your proof goal:

Choose any $$\epsilon>0$$ (this choice is out of your control, your proof must work for any choice). You want to produce an $$N\in\mathbb{N}$$ such that $$\{y_n\}$$ satisfies the definition.

First, pick some $$K>0$$ (you're allowed to do this since we know that $$x_n\rightarrow 0$$) such that if $$x_n>K$$ then $$y_n<\epsilon$$. This $$K$$ will be expressed algebraically in terms of $$\epsilon$$. (EDIT: I think you responded to another post very recently with this choice of $$K$$ - use that one.) Then, if you take some $$N_x$$ such that $$x_n>K$$ for all $$n\geq N_x$$, then you'll be able to say something about $$N_x$$ and $$\{y_n\}$$ similarly.